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\pub{2009}{1}{3}{1}
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\topic{Lecture 6 \\Differential Calculus-I\\ \scriptsize Partial Differentiation:Change of Variables (22 Sep 2009)}
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Let us consider
\[z=f(x,y)\]
If change in $x$ and $y$ variables be $\delta x$ and $\delta y$  respectively. Then change is $z$, $\delta z$ is
\[z+\delta z = f(x+\delta x, y+\delta y)\]
which gives
\[\delta z = f(x+\delta x, y+\delta y)-f(x,y)\]
Further, we may rearrange this as
\[\delta z = \frac{[f(x+\delta x, y+\delta y)-f(x, y+\delta y)]}{\delta x}\delta x+\frac{[f(x, y+\delta y)-f(x,y)]}{\delta y}\delta y\]
As $\delta x \rightarrow 0$ and $\delta y \rightarrow 0$, we get
\begin{equation}
dz = f_xdx+f_ydy
\end{equation}
\subsection*{Total Differential Coefficient}
If $z=f(x,y)$, where $x=\phi(t)$ and $y=\psi(t)$, then $z$ is called composite function of $t$. We can find Total Differential Coefficient\footnote{It can also be computed by substituting value of $x$ and $y$ in z, thus $z$ becomes function of $t$, which can be directly diffrentiated with respect to $t$.} $dz/dt$ of $z$ with respect to $t$ as
\begin{equation}
\frac{dz}{dt} = f_x\frac{dx}{dt}+f_y\frac{dy}{dt}
\end{equation}
\subsection*{Change of Variables}
If $z=f(x,y)$, where $x=\phi(r,s)$ and $y=\psi(r,s)$, we frequently need to change expression involving $x$, $y$ in to expression involving $r$, $s$, then we may use following relations 
\begin{equation}
\frac{\partial z}{\partial r} = f_x\frac{\partial x}{\partial r}+f_y\frac{\partial y}{\partial r}
\end{equation}
and
\begin{equation}
\frac{\partial z}{\partial s} = f_x\frac{\partial x}{\partial s}+f_y\frac{\partial y}{\partial s}
\end{equation}
%EXAMPLE
\begin{example}
If $u=x^3+y^3$, where $x=a \cos t$ and $y=b \sin t$, find $\frac{du}{dt}$
\end{example}
Here, $u=x^3+y^3 \Rightarrow u_x=3x^2,~~u_y=3y^2$,


Also $x=a \cos t$ and $y=b \sin t$ $\Rightarrow$ $\frac{dx}{dt} = -a \sin t$ and $\frac{dy}{dt} = b \cos t$ Respectively.
Now, using
\[\frac{du}{dt} = u_x\frac{dx}{dt}+u_y\frac{dy}{dt}\]
we get
\[\frac{du}{dt} = (3x^2)(-a \sin t)+ (3y^2)(b \cos t)\]
Change $x$ and $y$ in to $t$, $\frac{du}{dt} = -3a^3 \cos^2 t \sin t+ 3b^3 \sin^3 t \cos t)$

\begin{example}
If $u=u(x-y,y-z,z-x)$, show that $\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} + \frac{\partial u}{\partial z} = 0$
\end{example}
Let $r=x-y$, $s=y-z$, $t=z-x$, thus $u=u(r,s,t)$
\[\frac{\partial u}{\partial x} = u_r\frac{\partial r}{\partial x}+ u_s\frac{\partial s}{\partial x} + u_t\frac{\partial t}{\partial x} = u_r + 0 + -u_t\]
\[\frac{\partial u}{\partial y} = u_r\frac{\partial r}{\partial y}+ u_s\frac{\partial s}{\partial y} + u_t\frac{\partial t}{\partial y} = -u_r + u_s + 0\]
\[\frac{\partial u}{\partial z} = u_r\frac{\partial r}{\partial z}+ u_s\frac{\partial s}{\partial z} + u_t\frac{\partial t}{\partial z} = 0 + -u_s + u_t\]
On adding all these, we get
\[\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} + \frac{\partial u}{\partial z} = 0\]
\section*{Problems}
\begin{enumerate}
\item  If $u=f(x,y)$, where $x=e^r \cos \theta$ and $y=e^r \sin \theta$, then show that \[y\frac{\partial u}{\partial r} + x\frac{\partial u}{\partial \theta} = e^{2u}\frac{\partial u}{\partial y}\]
\item  If $u=f(x,y)$, where $x=r \cos \theta$ and $y=r \sin \theta$, then show that \[u_r^2 + \frac{1}{r^2}w_{\theta}^2 = f_x^2 + f_y^2\]
\item  If $u=u(\frac{y-x}{xy},\frac{z-x}{xz})$, show that \[x^2\frac{\partial u}{\partial x} + y^2\frac{\partial u}{\partial y} + z^2\frac{\partial u}{\partial z} = 0\]
\item  Transform the laplace equation (two variables) in to polar form.
\item  Prove that $u_{xx} + u_{yy}=u_{ss} + u_{tt} $, where
\[x=s \cos  \alpha - t \sin \alpha, ~~~y =s \sin \alpha + t \cos \alpha\]
\item  If $u=f(r)$, and $x=r \cos \theta$ and $y=r \sin \theta$, prove that
\[\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} =f''(r)+\frac{1}{r}f'(r)\]
\item  If $z$ is a function of $x$ and $y$, and $u$, $v$ be two other variables such that
$u=lx+my$, $v=ly-mx$, then show that
\[\frac{\partial^2 z}{\partial x^2} + \frac{\partial^2 z}{\partial y^2}=(l^2+m^2)\left(\frac{\partial^2 z}{\partial u^2} + \frac{\partial^2 z}{\partial v^2}\right)\]
\item  If $u \equiv x^2+y^2+z^2-2xyz =1$, show that
\[\frac{dx}{\sqrt{1-x^2}} + \frac{dy}{\sqrt{1-y^2}} + \frac{dz}{\sqrt{1-z^2}}=0 \]
\item  If $z=uv$ and $u^2+v^2-x-y=0$, $u^2-v^2+3x+y=0$, find \pd{z}{x}
\end{enumerate}

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